Question

Fluid used in the column is mercury. The pressure of the gas in the cylinder is :
Given : Atmospheric pressure $=1.01\times {10}^5{\text{N/m}}^2,g=10{\text{m/s}}^2$ and density of mercury $=13600{\text{kg/m}}^3$.

• A. $1.28\times {10}^5{\text{N/m}}^2$
• B. $1.31\times {10}^5{\text{N/m}}^2$
• C. $1.11\times {10}^5{\text{N/m}}^2$
• D. $1.23\times {10}^5{\text{N/m}}^2$

Answer 1

The correct option is D $1.23\times {10}^5{\text{N/m}}^2$

Let the pressure of the gas in the cylinder be $P$.


Pressure of the gas remains same throughout the volume occupied by the gas, as density of the gas is very very low. So we neglect the hydrostatic pressure due to the gas column.

Equating the pressure in the same liquid at same elevation,
$P=P_A$
$\Rightarrow P=P_0+\rho gh$,
Here, $h=0.20-0.04=0.16\text{cm}$
$\Rightarrow P=1.01\times {10}^5+13600\times 10\times 0.16$
$\therefore P=1.23\times {10}^5{\text{N/m}}^2$